Newton-Cotes says "pick evenly spaced points in the interval, draw the interpolating polynomial of minimal degree through them, and integrate the polynomial." From the Lagrange interpolation theorem, a Newton-Cotes rule on $n$ nodes is exact for polynomials of degree at most $n-1$. It is not exact for polynomials of any higher degree, because one can add a polynomial of degree $n$ which vanishes at the $n$ nodes and takes on any desired value at an additional point, and the Newton-Cotes rule will not "detect" this change.
Gaussian quadrature says "find a rule that lets us exactly integrate polynomials of as large a degree as possible." This winds up being degree $2n-1$, since we have control of $2n$ parameters in specifying a quadrature rule at $n$ nodes, and the parameters are independent.
Analysis of either method is generally inspired by the Weierstrass approximation theorem, which generates the following type of argument. Here $I$ denotes exact integration, $Q_n$ denotes quadrature integration with a rule on $n$ nodes, and $p$ is a polynomial.
$$| I(f) - Q_n(f) | \leq | I(f) - I(p) | + | I(p) - Q_n(p) | + | Q_n(p) - Q_n(f) | $$
The Weierstrass theorem and the fact that $I$ is a bounded linear functional (in particular, $\| I \| = I(1)$) allow us to make the first term small, possibly at the expense of $p$ having a large degree.
Once the degree of $p$ is fixed, choosing $n$ large enough enables us to make the second term zero, since both Newton-Cotes and Gaussian quadrature can be made exact for polynomials by taking enough nodes.
The problem for the numerical analyst is to control the last term, when $\| p - f \|$ is small and $n$ might be arbitrarily large.
Now $Q_n$, like $I$, is linear, so we can write
$$Q_n(p) - Q_n(f) = Q_n(p-f).$$
Also, $Q_n$ is bounded:
$$|Q_n(f)| \leq \| f \| \sum_{k=1}^n |w_k|,$$
where $w_k$ are the weights. Hence
$$\| Q_n \| \leq \sum_{k=1}^n |w_k|.$$
This inequality is actually an equality, as we can check by integrating $f$ such that $\| f \|=1$ and $f(x_k)=\text{sign}(w_k)$.
The fact that the $Q_n$ are each bounded is not good enough. Since simultaneously controlling the first two terms takes away our control over $n$, we need $Q_n$ to be uniformly bounded.
It turns out that Newton-Cotes rules do not have $Q_n$ uniformly bounded, whereas Gaussian quadrature does. The idea is to recognize that, because these methods are exact for polynomials of degree $0$, $\sum_{k=1}^n w_k = I(1)$. If the $w_k$ are all positive, then $\| Q_n \| = \sum_{k=1}^n |w_k| = I(1)$, which is independent of $n$, as we want.
One can prove from the property of being exact for polynomials of degree $2n-1$ that all of the weights in Gaussian quadrature are positive. (The trick is to integrate $\prod_{k=1,k \neq j}^n (x-x_k)^2$ for each $j$.) Consequently Gaussian quadrature is convergent. In particular, if there exists $p$ such that $\| p - f \| < \varepsilon/(2I(1))$ and the degree of $p$ is at most $m$, then $|I(f)-Q_n(f)|<\varepsilon$ provided $n \geq (m+1)/2$.
The weights in Newton-Cotes quadrature are not positive. Moreover, $\| Q_n \|$ is unbounded for Newton-Cotes rules. This means that if $f=p+g$, where $p$ is a polynomial of degree at most $n-1$ and $g$ is a continuous function with $g(x_k)=\| g \| \text{sign}(w_k)$, then $|Q_n(p-f)|=\| Q_n \| \| g \| \gg \| g \|$.
Consequently Newton-Cotes rules are not convergent for all continuous $f$. Very "nice" functions $f$ will converge with Newton-Cotes rules, but Runge's well-known example with $1/(x^2+1)$ shows that the number of derivatives is not what "nice" means. (The answer turns out to be that the Fourier coefficients decay sufficiently fast.)
Best Answer
Actually, your points are $x_1=a+1\cdot\frac{b-a}3=\frac{2a+b}3$ and $x_2=a+2\cdot\frac{b-a}3=\frac{a+2b}3$
Now determine weights $w_1,w_2$ so that $$ w_1f(x_1)+w_2f(x_2)=\int_a^b f(x)\,dx $$ for $f(x)=1$ and $f(x)=x$, and if possible, also for $f(x)=x^2$.